The lifespans of seals in a particular zoo are normally distributed. The average seal lives $13.2$ years; the standard deviation is $2.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living less than $15.9$ years.
Solution: $13.2$ $10.5$ $15.9$ $7.8$ $18.6$ $5.1$ $21.3$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $13.2$ years. We know the standard deviation is $2.7$ years, so one standard deviation below the mean is $10.5$ years and one standard deviation above the mean is $15.9$ years. Two standard deviations below the mean is $7.8$ years and two standard deviations above the mean is $18.6$ years. Three standard deviations below the mean is $5.1$ years and three standard deviations above the mean is $21.3$ years. We are interested in the probability of a seal living less than $15.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the seals will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $10.5$ years and the other half $({16\%})$ will live longer than $15.9$ years. The probability of a particular seal living less than $15.9$ years is ${68\%} + {16\%}$, or $84\%$.